Show that the middle term in the expansion of(1+x)raise to power 2n is = 1.3.5.......(2n-1) . 2n.xraise to power n upon n! , where nis a +ve integer.
Total no of terms are (2n+1) which are odd (right?), so middle term will be (2n+1 +1) /2 th term i.e. (n+1)th term
Now middle term i.e. (n+1) th term of (1+x)2n = 2nCn (x)n, = !2n /(!n *!n) = 184.108.40.206.......(2n-1) (2n)xn/ (!n * !n)
(remember 1.2.3.....(2n) are total '2n' terms out of which 'n' will be even (i.e. 2.4.6.....upto (2n) ) and 'n' are odd (i.e. 220.127.116.11..upto (2n-1) ).
18.104.22.168.......(2n-1) (2n)/ (!n * !n) = 22.214.171.124...(2n-1) * (2.4.6.....2n)xn /!n *!n
Taking out '2' each from these 'n' 'even' termswe get 1.3.5...(2n-1) (2n)(126.96.36.199.5.....n) xn/!n *!n,
or ,1.3.5...(2n-1) (2n) !n/(!n*!n) which gives 1.3.5...(2n-1) (2n)xn/ !n